# Problem 14.5a2

Counting states in heterotic SO(32) string theory | A First Course in String Theory

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(a) Consider the left NS’ sector. Write the precise mass-squared formula with normal-ordered oscillators and the appropriate normal-ordering constant.

~~~

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$\displaystyle{\alpha' M_L^2 = \frac{1}{2} \sum_{n \ne 0} \bar \alpha_{-n}^I \bar \alpha_n^I + \frac{1}{2} \sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}$

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— This answer is my guess. —

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Equation at Problem 14.5:

$\displaystyle{\alpha' M_L^2}$

$\displaystyle{= \frac{1}{2} \sum_{n \ne 0} \bar \alpha_{-n}^I \bar \alpha_n^I + \frac{1}{2} \sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}$

$\displaystyle{= \frac{-1}{8} + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I + \frac{1}{2} \sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}$

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$\displaystyle{\sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}$
$\displaystyle{= \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \left[ 2 \lambda_{-r}^A \lambda_r^A - 1 \right]}$

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Equation (13.116):

$\displaystyle{\sum_{k \in \mathbf{Z}^+_{\text{odd}}} k = \frac{1}{12}}$

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\displaystyle{\begin{aligned} &\sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A \\ &= \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \left[ 2 \lambda_{-r}^A \lambda_r^A - 1 \right] \\ &= - \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r + 2 \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \lambda_{-r}^A \lambda_r^A \\ &= - \frac{1}{2} \sum_{r = 1, 3, ...} r + 2 \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \lambda_{-r}^A \lambda_r^A \\ &= - \frac{1}{24} + 2 \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \lambda_{-r}^A \lambda_r^A \end{aligned}}

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\displaystyle{ \begin{aligned} \alpha' M_L^2 &= \frac{-7}{48} + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I + \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \lambda_{-r}^A \lambda_r^A \\ \end{aligned}}

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If we define $N^\perp$ in the way similar to equation (14.37), we have

\displaystyle{ \begin{aligned} \alpha' M_L^2 &= \frac{-7}{48} + N^\perp \\ \end{aligned}}

— This answer is my guess. —

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— Me@2018-09-01 06:05:29 AM

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# Problem 14.5a1

Counting states in heterotic $SO(32)$ string theory | A First Course in String Theory

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(a) Consider the left NS’ sector. Write the precise mass-squared formula with normal-ordered oscillators and the appropriate normal-ordering constant.

~~~

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$\displaystyle{\alpha' M_L^2 = \frac{1}{2} \sum_{n \ne 0} \bar \alpha_{-n}^I \bar \alpha_n^I + \frac{1}{2} \sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}$

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What is normal-ordering?

Put all the creation operators on the left.

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What for?

p.251 “It is useful to work with normal-ordered operators since they act in a simple manner on the vacuum state. We cannot use operators that do not have a well defined action on the vacuum state.”

“The vacuum expectation value of a normal ordered product of creation and annihilation operators is zero. This is because, denoting the vacuum state by $|0\rangle$, the creation and annihilation operators satisfy”

$\displaystyle{\langle 0 | \hat{a}^\dagger = 0 \qquad \textrm{and} \qquad \hat{a} |0\rangle = 0}$

— Wikipedia on Normal order

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— This answer is my guess. —

$\displaystyle{\sum_{n \ne 0} \bar \alpha_{-n}^I \bar \alpha_n^I}$

$\displaystyle{= \sum_{n \in \mathbf{Z}^-} \bar \alpha_{-n}^I \bar \alpha_n^I + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_n^I}$

$\displaystyle{= \sum_{n \in \mathbf{Z}^+} \bar \alpha_{n}^I \bar \alpha_{-n}^I + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_n^I}$

$\displaystyle{= \sum_{n \in \mathbf{Z}^+} \left[ \bar \alpha_{n}^I \bar \alpha_{-n}^I - \bar \alpha_{-n}^I \bar \alpha_{n}^I + \bar \alpha_{-n}^I \bar \alpha_{n}^I \right] + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_n^I}$

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$\displaystyle{= \sum_{n \in \mathbf{Z}^+} \left[ \bar \alpha_{n}^I, \bar \alpha_{-n}^I \right] + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_n^I}$

$= \displaystyle{\sum_{n \in \mathbf{Z}^+} n \eta^{II} + 2 \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I}$

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c.f. p.251:

$\displaystyle{\sum_{n \ne 0} \bar \alpha_{-n}^I \bar \alpha_n^I}$

$\displaystyle{= \sum_{n \in \mathbf{Z}^+} n \eta^{II} + 2 \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I}$

$\displaystyle{= \frac{-1}{12} (D - 2) + 2 \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I}$

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Equation at Problem 14.5:

$\displaystyle{\alpha' M_L^2}$

$\displaystyle{= \frac{1}{2} \sum_{n \ne 0} \bar \alpha_{-n}^I \bar \alpha_n^I + \frac{1}{2} \sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}$

$\displaystyle{= \frac{1}{2} \left[ \frac{-1}{12} (D - 2) + 2 \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I \right] + \frac{1}{2} \sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}$

$\displaystyle{= \frac{-1}{24} (D - 2) + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I + \frac{1}{2} \sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}$

$\displaystyle{= \frac{-1}{8} + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I + \frac{1}{2} \sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}$

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$D = 10$

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$\displaystyle{\sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}$

$\displaystyle{= \sum_{r = - \frac{1}{2}, - \frac{3}{2}, ...} r \lambda_{-r}^A \lambda_r^A + \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \lambda_{-r}^A \lambda_r^A}$

$\displaystyle{= \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} (-r) \lambda_{r}^A \lambda_{-r}^A + \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \lambda_{-r}^A \lambda_r^A}$

$\displaystyle{= \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \left[ (-1) \lambda_{r}^A \lambda_{-r}^A + \lambda_{-r}^A \lambda_r^A \right]}$

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$\displaystyle{= \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \left[ (-1) \lambda_{r}^A \lambda_{-r}^A + \lambda_{-r}^A \lambda_r^A \right]}$

$\displaystyle{= \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \left[ \lambda_{-r}^A, \lambda_r^A \right]}$

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Equation (14.29):

$\displaystyle{\left\{ b_r^I, b_s^J \right\} = \delta_{r+s, 0} \delta^{IJ}}$

$\displaystyle{b_r^I b_s^J = - b_s^I b_r^J + \delta_{r+s, 0} \delta^{IJ}}$

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$\displaystyle{\sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}$

$\displaystyle{= \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \left[ (-1) \lambda_{r}^A \lambda_{-r}^A + \lambda_{-r}^A \lambda_r^A \right]}$

$\displaystyle{= \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \left[ (-1) \left( - \lambda_{-r}^A \lambda_r^A + \delta_{r-r, 0} \delta^{AA} \right) + \lambda_{-r}^A \lambda_r^A \right]}$

$\displaystyle{= \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \left[ \lambda_{-r}^A \lambda_r^A + \lambda_{-r}^A \lambda_r^A - 1 \right]}$

$\displaystyle{= \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \left[ 2 \lambda_{-r}^A \lambda_r^A - 1 \right]}$

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$\displaystyle{\sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}$

$\displaystyle{= - \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r + \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \left[ b_{-r}^A b_r^A + \lambda_{-r}^A \lambda_r^A \right]}$

$\displaystyle{= - \frac{1}{2} \sum_{r = 1, 3, ...} r + \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \left[ b_{-r}^A b_r^A + \lambda_{-r}^A \lambda_r^A \right]}$

$\displaystyle{= \left[ - \frac{1}{24} + \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \left( b_{-r}^A b_r^A + \lambda_{-r}^A \lambda_r^A \right) \right]}$

— This answer is my guess. —

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— Me@2018-08-06 10:23:48 PM

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# Problem 14.4b2

Closed string degeneracies | A First Course in String Theory

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(b) State the values of $\alpha' M^2$ and give the separate degeneracies of bosons and fermions for the first five mass levels of the type IIA closed superstrings. Would the answer have the different for type IIB?

~~~

Type IIB closed superstrings

Equation (14.85)

$(NS+, NS+), (NS+, R-), (R-, NS+), (R-, R-)$

— Me@2015.09.16 06:08 AM: Should be the same. But I am not sure whether I have missed something.

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$f_{NS+}(x) = 8 + 128 \, x + 1152 \, x^{2} + 7680 \, x^{3} + 42112 \, x^{4} + ...$

$f_{R-}(x) = 8 + 128 x + 1152 x^{2} + 7680 x^{3} + 42112 x^{4} + ...$

$f_{NS-}(x) = \frac{1}{\sqrt{x}} + 36 \sqrt{x} + 402 x^{\frac{3}{2}} + 3064 x^{\frac{5}{2}} + ...$

$f_{R+}(x) = 8 + 128 x + 1152 x^{2} + 7680 x^{3} + 42112 x^{4} + ...$

— Me@2018-07-14 09:41:10 PM

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# Quick Calculation 14.8.2

A First Course in String Theory

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What sector(s) can be combined with a left-moving NS- to form a consistent closed string sector?

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There are no mass levels in NS+, R+, or R- that can match those in NS-. So NS- can be paired only with NS-:

$(NS-, NS-)$

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$f_{NS} (x)$
$= \frac{1}{\sqrt{x}} \prod_{n=1}^\infty \left( \frac{1+x^{n-\frac{1}{2}}}{1-x^n} \right)^8$
$= \frac{1}{\sqrt{x}} g_{NS}(x)$
$= \frac{1}{\sqrt{x}} + 8 + 36 \sqrt{x} + 128 x + 402 x \sqrt{x} + 1152 x^2 + ...$

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$g (\sqrt{x})$
$= \prod_{n=1}^\infty \left( \frac{1+x^{n-\frac{1}{2}}}{1-x^n} \right)^8$
$= 1 + 8 \, \sqrt{x} + 36 \, x + 128 \, x^{\frac{3}{2}} + 402 \, x^{2} + 1152 \, x^{\frac{5}{2}} + 3064 \, x^{3} + ...$

$g (-\sqrt{x})$
$= \prod_{n=1}^\infty \left( \frac{1-x^{n-\frac{1}{2}}}{1-x^n} \right)^8$
$= 1 -8 \, \sqrt{x} + 36 \, x -128 \, x^{\frac{3}{2}} + 402 \, x^{2} -1152 \, x^{\frac{5}{2}} + 3064 \, x^{3} + ...$

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$g (\sqrt{x}) + g (-\sqrt{x})$
$= 2(1 + 36 x + 402 x^{2} + 3064 x^{3} + ...)$

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$f_{NS-}(x)$
$= \frac{1}{2 \sqrt{x}} \left[ g (\sqrt{x}) + g (-\sqrt{x}) \right]$
$= \frac{1}{2 \sqrt{x}} \left[ \prod_{n=1}^\infty \left( \frac{1+x^{n-\frac{1}{2}}}{1-x^n} \right)^8 + \prod_{n=1}^\infty \left( \frac{1-x^{n-\frac{1}{2}}}{1-x^n} \right)^8 \right]$
$= \frac{1}{2 \sqrt{x}} \left[ 2(1 + 36 \, x + 402 \, x^{2} + 3064 \, x^{3} + ...) \right]$
$= \frac{1}{\sqrt{x}} + 36 \sqrt{x} + 402 x^{\frac{3}{2}} + 3064 x^{\frac{5}{2}} + ...$

— Me@2018-06-26 07:36:41 PM

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# Problem 14.4b1.4

Closed string degeneracies | A First Course in String Theory

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What is the meaning of “With $a = 1, ..., 8$ and $\bar b = \bar 1, ..., \bar 8$, …”?

— Me@2015.09.14 12:11 PM

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p.315 “Explicitly, the eight states $| R_a \rangle, a = 1, 2, ..., 8$, with an even number of creation operators are … ”

p.316 “The eight states $|R_{\bar{a}} \rangle, \bar a = \bar 1, \bar 2, ..., \bar 8$, with an odd number of creation operators are … ”

— Me@2018-05-24 11:41:34 AM

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# Problem 14.4b1.3

Closed string degeneracies | A First Course in String Theory

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(b) State the values of $\alpha' M^2$ and give the separate degeneracies of bosons and fermions for the first five mass levels of the type IIA closed superstrings. Would the answer have the different for type IIB?

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— This answer is my guess. —

Since for NS, the first 5 levels’ degeneracies are 8, 36, 128, 402, 1152, the degeneracies of (NS, NS) are $8^2, 36^2, 128^2, 402^2, 1152^2$.

This is incorrect, for there are no (NS, NS) states. Instead, you should consider (NS+, NS+).

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Since for NS+, the first 5 levels’ degeneracies are 8, 128, 1152, 7680, 42112, the degeneracies of (NS+, NS+) are $8^2, 128^2, 1152^2, 7680^2, 42112^2$.

p.317 Consider the relationship of the degeneracy of R+ and that of R-:

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How about the first 5 levels of R+?

The degeneracies are the same as those of R-.

p.317 Equation (14.54) “The appearance of an equal number of bosonic and fermionic states at every mass level is a signal of supersymmetry. This is, however, supersymmetry on the world-sheet.”

Equation (14.71):

$f_{R-}(x) = 8 + 128 x + 1152 x^{2} + 7680 x^{3} + 42112 x^{4} + ...$

p.321 “Indeed, the integer mass-squared levels in the NS generating function (14.67) have degeneracies that match those of (14.71) for the R- sector.”

— This answer is my guess. —

— Me@2018-05-14 02:51:55 PM

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# Problem 14.4b1.2

Closed string degeneracies | A First Course in String Theory

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(b) State the values of $\alpha' M^2$ and give the separate degeneracies of bosons and fermions for the first five mass levels of the type IIA closed superstrings. Would the answer have the different for type IIB?

~~~

NS+ equations of (14.38):

 $\alpha'M^2=0,$ $~~N^\perp = \frac{1}{2}:$ $~~~~b_{-1/2}^I |NS \rangle \otimes |p^+, \vec p_T \rangle,$ $\alpha'M^2=1,$ $~~N^\perp = \frac{3}{2}:$ $~~~~\{ \alpha_{-1}^I b_{\frac{-1}{2}}^J, b_{\frac{-3}{2}}^I, b_{\frac{-1}{2}}^I b_{\frac{-1}{2}}^J b_{\frac{-1}{2}}^K \} |NS \rangle \otimes |p^+, \vec p_T \rangle,$ $\alpha'M^2=2,$ $~~N^\perp = \frac{5}{2}:$ $~~~~\{\alpha_{-2}^I b_{\frac{-1}{2}}^J, \alpha_{-1}^I \alpha_{-1}^J b_{\frac{-1}{2}}^K, \alpha_{-1}^I b_{\frac{-3}{2}}^J, \alpha_{-1}^I b_{\frac{-1}{2}}^J b_{\frac{-1}{2}}^K b_{\frac{-1}{2}}^M, ...\}$ $\{ ..., b_{\frac{-5}{2}}^I, b_{\frac{-3}{2}}^I b_{\frac{-1}{2}}^J b_{\frac{-1}{2}}^K, b_{\frac{-1}{2}}^I b_{\frac{-1}{2}}^J b_{\frac{-1}{2}}^K b_{\frac{-1}{2}}^M b_{\frac{-1}{2}}^N \} |NS \rangle \otimes |p^+, \vec p_T \rangle,$ … … …

For $N^\perp = \frac{5}{2}$, the number of states is

$8^2 + \left[ \frac{(8)(7)}{2!} + 8 \right] (8) + 8^2$
$+ 8 + \frac{(8)(7)(6)}{3!} + 8 + (8) \left[ \frac{(8)(7)}{2!} \right] + \frac{(8)(7)(6)(5)(4)}{5!}$
$= 1152$

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Since $\alpha' M^2 = N^\perp - \frac{1}{2}$, when $N^\perp = \frac{5}{2}$, $\alpha' M^2 = 2$.

Equation (14.67):

$f_{NS} (x) = \frac{1}{\sqrt{x}} + 8 + 36 \sqrt{x} + 128 x + 402 x \sqrt{x} + 1152 x^2 + ...$

Equation (14.66):

$f_{NS} (x) = \frac{1}{\sqrt{x}} \prod_{n=1}^\infty \left( \frac{1+x^{n-\frac{1}{2}}}{1-x^n} \right)^8$

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p.321

If we take $f_{NS} (x)$ in (14.66) and change the sign inside each factor in the numerator

Equation (14.72):

$\frac{1}{\sqrt{x}} \prod_{n=1}^\infty \left( \frac{1-x^{n-\frac{1}{2}}}{1-x^n} \right)^8$

the _only_ effect is changing the sign of each term in the generating function whose states arise with an odd number of fermions.

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$\prod_{n=1}^\infty \left( \frac{1}{1-x^n} \right)^8$ is the boson contribution.

$\prod_{n=1}^\infty \left( 1+x^{n-\frac{1}{2}} \right)^8$ is the fermion contribution.

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Turning Equation (14.67) into

$f_{NS?} (x)$
$= - \frac{1}{\sqrt{x}} + 8 - 36 \sqrt{x} + 128 x - 402 x \sqrt{x} + 1152 x^2 - ...$

is equivalent to turning all $\sqrt{x}$ into $- \sqrt{x}$:

$f_{NS?} (x)$
$= \frac{1}{\sqrt{x}} \prod_{n=1}^\infty \left( \frac{1-x^{n-\frac{1}{2}}}{1-x^n} \right)^8$

Me@2015.08.29 12:49 PM: Somehow, $\sqrt{x}$ represents “contribution from fermions”.

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Me@2015.08.29 12:50 PM: If you still cannot understand, try replace all $\sqrt{x}$ with $y$.

$f_{NS+} (x) = \frac{1}{2} \left( f_{NS} - f_{NS?} \right)$

$f_{NS} (x)$
$= \frac{1}{\sqrt{x}} + 8 + 36 \sqrt{x} + 128 x + 402 x \sqrt{x} + 1152 x^2 + ...$

$f_{NS?} (x)$
$= - \frac{1}{\sqrt{x}} + 8 - 36 \sqrt{x} + 128 x - 402 x \sqrt{x} + 1152 x^2 - ...$

$f_{NS+} (x)$
$= \frac{1}{2} \left( f_{NS} - f_{NS?} \right)$
$= \frac{1}{\sqrt{x}} + 36 \sqrt{x} + 402 x \sqrt{x} + ...$

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It is _not_ correct. Just consider it as $\left(\sqrt{x} \to -\sqrt{x} \right)$ is not correct, since the beginning factor $\frac{1}{\sqrt{x}}$ is not considered yet.

Instead, we should present in the following way:

$f_{NS} (x)$
$= \frac{1}{\sqrt{x}} \prod_{n=1}^\infty \left( \frac{1+x^{n-\frac{1}{2}}}{1-x^n} \right)^8$
$= \frac{1}{\sqrt{x}} g_{NS}(x)$
$= \frac{1}{\sqrt{x}} + 8 + 36 \sqrt{x} + 128 x + 402 x \sqrt{x} + 1152 x^2 + ...$

$g (\sqrt{x})$
$= \prod_{n=1}^\infty \left( \frac{1+x^{n-\frac{1}{2}}}{1-x^n} \right)^8$
$= 1 + 8 \, \sqrt{x} + 36 \, x + 128 \, x^{\frac{3}{2}} + 402 \, x^{2} + 1152 \, x^{\frac{5}{2}} + 3064 \, x^{3} + ...$

$g (-\sqrt{x})$
$= \prod_{n=1}^\infty \left( \frac{1-x^{n-\frac{1}{2}}}{1-x^n} \right)^8$
$= 1 -8 \, \sqrt{x} + 36 \, x -128 \, x^{\frac{3}{2}} + 402 \, x^{2} -1152 \, x^{\frac{5}{2}} + 3064 \, x^{3} + ...$

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$g (\sqrt{x}) - g (-\sqrt{x})$
$= \prod_{n=1}^\infty \left( \frac{1+x^{n-\frac{1}{2}}}{1-x^n} \right)^8 - \prod_{n=1}^\infty \left( \frac{1-x^{n-\frac{1}{2}}}{1-x^n} \right)^8$
$= 16 \, \sqrt{x} + 256 \, x^{\frac{3}{2}} + 2304 \, x^{\frac{5}{2}} + 15360 \, x^{\frac{7}{2}} + ...$

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$f_{NS+}(x) = \frac{1}{2 \sqrt{x}} \left[ g (\sqrt{x}) - g (-\sqrt{x}) \right]$
$= \frac{1}{2 \sqrt{x}} \left[ \prod_{n=1}^\infty \left( \frac{1+x^{n-\frac{1}{2}}}{1-x^n} \right)^8 - \prod_{n=1}^\infty \left( \frac{1-x^{n-\frac{1}{2}}}{1-x^n} \right)^8 \right]$
$= \frac{1}{2 \sqrt{x}} \left[ 16 \, \sqrt{x} + 256 \, x^{\frac{3}{2}} + 2304 \, x^{\frac{5}{2}} + 15360 \, x^{\frac{7}{2}} + ... \right]$
$= 8 + 128 \, x + 1152 \, x^{2} + 7680 \, x^{3} + 42112 \, x^{4} + ...$

— Me@2018-05-08 08:50:32 PM

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# Problem 14.4b1.1

Closed string degeneracies | A First Course in String Theory

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(b) State the values of $\alpha' M^2$ and give the separate degeneracies of bosons and fermions for the first five mass levels of the type IIA closed superstrings. Would the answer have the different for type IIB?

~~~

Type IIA closed superstrings

p.322

In closed superstring theories spacetime bosons arise from the (NS, NS) sector and also from the (R, R) sector, since this sector is “doubly” fermionic. The spacetime fermions arise from the (NS, R) and (R, NS) sectors.

p.322 $\alpha' M_L^2 = \alpha' M_R^2$

 $\frac{1}{2} \alpha' M^2 =$ $\alpha' M_L^2 + \alpha' M_R^2$ $\alpha' M^2 =$ $2 \left( \alpha' M_L^2 + \alpha' M_R^2 \right)$ $=$ $4 \alpha' M_L^2$

Equation (14.77):

$\text{type IIA}:~~~(NS+, NS+), ~(NS+, R+),~ (R-, NS+), ~ (R-, R+)$

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What is the difference of the meanings of R+ and R-?

$R+$ states are world-sheet bosonic states.

p.316

It thus follows that all eight $| R_a \rangle$ states are fermionic and all $| R_{\bar a} \rangle$ are bosonic.

Be careful:

Here, “fermionic”/”bosonic” refers to the world-sheet fermions/bosons, not the spacetime ones.

p.320

Identifying $| R_a \rangle$ as spacetime fermions and $| R_{\bar a} \rangle$ as spacetime bosons is not an alternative either, since spacetime bosons cannot carry a spinor index.

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How come the R+ cannot be the left-moving part?

p.320

A strategy then emerges. Since all states in the R sector have a spinor index, we will only attempt to get spacetime fermions from this sector. We also recognize that all fermions must arise from states with the same value of $(-1)^F$.

Me@2015.09.11 10:36 AM: In other words, it is a convention:

Following Gliozzi, Scherk, and Olive (GSO) we proceed to truncate the Ramond sector down to the set of states with $(-1)^F = -1$.

— Me@2018-05-01 05:59:53 PM

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# Problem 14.4a4

Closed string degeneracies | A First Course in String Theory

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(a) State the values of $\alpha' M^2$ and give the degeneracies for the first five mass levels of the closed bosonic string theory.

~~~

 $|$ $~\text{Number of states}$ ${a_1^I}^\dagger {a_1^J}^\dagger \bar a_1^{K\dagger} \bar a_1^{L\dagger} | p^+, \vec p_T \rangle~|$ $~\left[ \frac{(D-2)(D-1)}{2} \right]^2$ ${a_1^I}^\dagger {a_1^J}^\dagger \bar a_2^{K\dagger} | p^+, \vec p_T \rangle~|$ $~\left[ \frac{(D-2)(D-1)}{2} \right](D-2)$ ${a_2^I}^\dagger \bar a_1^{K\dagger} \bar a_1^{L\dagger} | p^+, \vec p_T \rangle~|$ $~(D-2)\left[ \frac{(D-2)(D-1)}{2} \right]$ ${a_2^I}^\dagger \bar a_2^{K\dagger} | p^+, \vec p_T \rangle~|$ $~(D-2)^2$

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Can we create a formula for the number of states?

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$\left[ \frac{(D-2)(D-1)}{2} \right]^2 + \left[ \frac{(D-2)(D-1)}{2} \right](D-2) + (D-2)\left[ \frac{(D-2)(D-1)}{2} \right] + (D-2)^2$
$= ...$
$= (D-2)^2\left\{ \frac{1}{2} \frac{(D-1)^2}{2} + D \right\}$
$= 104976$
$= 324^2$
$= \left[ \frac{(D-2)(D-1)}{2} + (D-2) \right]^2$

The result is the same as the square of the coefficients of $x$ in Equation (14.63) on page 318.

 $\frac{1}{2} \alpha' M^2~|$ $N~|$ $~\bar N~$ $|~\text{Number of states}$ $-2~|$ $0~|$ $~0~$ $|~1$ $0~|$ $1~|$ $~1~$ $|~(D-2)^2$ $2~|$ $2~|$ $~2~$ $|~(D-2)^2\left\{ \frac{1}{2} \frac{(D-1)^2}{2} + D \right\}$ $4~|$ $3~|$ $~3~$ $|~3200^2$ $8~|$ $4~|$ $~4~$ $|~25650^2$

— Me@2018-04-25 05:13:04 PM

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# Problem 14.4a3

Closed string degeneracies | A First Course in String Theory

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(a) State the values of $\alpha' M^2$ and give the degeneracies for the first five mass levels of the closed bosonic string theory.

~~~

But for non-massless states, this probably is not true anymore:

 $|$ $~\text{Number of states}$ ${a_1^I}^\dagger {a_1^J}^\dagger \bar a_1^{K\dagger} \bar a_1^{L\dagger} | p^+, \vec p_T \rangle~|$ $~\left[ \frac{(D-2)(D-1)}{2} \right]^2$ ${a_1^I}^\dagger {a_1^J}^\dagger \bar a_2^{K\dagger} | p^+, \vec p_T \rangle~|$ $~\left[ \frac{(D-2)(D-1)}{2} \right](D-2)$ ${a_2^I}^\dagger \bar a_1^{K\dagger} \bar a_1^{L\dagger} | p^+, \vec p_T \rangle~|$ $~(D-2)\left[ \frac{(D-2)(D-1)}{2} \right]$ ${a_2^I}^\dagger \bar a_2^{K\dagger} | p^+, \vec p_T \rangle~|$ $(D-2)^2$

So the total number of states for $\frac{1}{2} \alpha' M^2 = 2$ ($N = \bar N = 2$) is

$\left[ \frac{(D-2)(D-1)}{2} \right]^2 + \left[ \frac{(D-2)(D-1)}{2} \right](D-2)$
$+ (D-2)\left[ \frac{(D-2)(D-1)}{2} \right] + (D-2)^2$

.

Should $D$ be 10 or 26?

p.324 “Out of 26 left-moving bosonic coordinates of the bosonic factor only ten of them are matched by the right-moving bosonic coordinates of the superstring factor.”

$D$ should be 26 for bosonic strings. So the total number of states is

 $\frac{1}{2} \alpha' M^2~|$ $~\text{Number of states}$ $2~|$ $~104976$

.
What does the difference of this part and Section 14.6 come from?

This part is for bosonic closed string, while Section 14.6 is for bosonic open string. There is no $\bar N$ to consider in Section 14.6.

p.290 “A basis vector $| \lambda, \bar \lambda \rangle$ belongs to the state space if and only if it satisfies the level-matching constraint”

$N^\perp = \bar N^\perp$.

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Can we create a formula for the number of states?

— Me@2018-04-23 11:31:16 AM

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# Quick Calculation 14.6b

A First Course in String Theory

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Construct explicitly all the states with $\alpha' M^2=2$ and count them, verifying that there are indeed a total of 3200 states. You may find the counting formula in Problem 12.11 useful.

~~~

Let $N(n, k) = {n + k - 1 \choose k - 1}$, the number of ways to put n indistinguishable balls into k boxes.

p.318 “For open bosonic strings $\alpha' M^2 = N^\perp - 1$, …”

.

When $\alpha' M^2 = 2, N^\perp = 3$, the cases are:

1. three $a_1^\dagger$‘s:

$N(3,24) = 2600$

2. one $a_2^\dagger$ and one $a_1^\dagger$:

$24 \times 24 = 576$

3. one $a_3^\dagger$:

24

.

Total number of possible states for $N^\perp = 3$ is 3200.

— Me@2018-04-20 02:45:35 PM

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# Problem 14.4a2

Closed string degeneracies | A First Course in String Theory

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(a) State the values of $\alpha' M^2$ and give the degeneracies for the first five mass levels of the closed bosonic string theory.

~~~

p.288 Equation (13.48):

$M^2 = \frac{2}{\alpha'} \left( N^\perp + \bar N^\perp - 2 \right)$

.

p.290 “A basis vector $| \lambda, \bar \lambda \rangle$ belongs to the state space _if and only if_ it satisfies the level-matching constraint”

$N^\perp = \bar N^\perp$

.

 $\frac{1}{2} \alpha' M^2~|$ $~\text{Number of states}$ $-2~|$ $~1$ $0~|$ $~(D-2)^2$

.

 $\frac{1}{2} \alpha' M^2~|$ $~\text{Number of states}$ $2~|$ $...$

${a_1^I}^\dagger {a_1^J}^\dagger \bar a_1^{K\dagger} \bar a_1^{L\dagger} | p^+, \vec p_T \rangle$
${a_1^I}^\dagger {a_1^J}^\dagger \bar a_2^{K\dagger} | p^+, \vec p_T \rangle$
${a_2^I}^\dagger \bar a_1^{K\dagger} \bar a_1^{L\dagger} | p^+, \vec p_T \rangle$

Assuming the signs of the wave functions do not matter:

 $|$ $|~\text{Number of states}$ ${a_1^I}^\dagger {a_1^J}^\dagger \bar a_1^{K\dagger} \bar a_1^{L\dagger}~|$ $\left[ \frac{(D-2)(D-1)}{2} \right]^2$ ${a_1^I}^\dagger {a_1^J}^\dagger \bar a_2^{K\dagger} | p^+, \vec p_T \rangle~|$ ${a_2^I}^\dagger \bar a_1^{K\dagger} \bar a_1^{L\dagger} | p^+, \vec p_T \rangle~|$

p.291 How come the number of the components of the matrix represents the number of states?

p.292 For massless states, we have only one ${a_1^{I}}^\dagger$ and $\bar a_1^{J\dagger}$, where $a$ and $\bar a$ cannot interchange. So $I$ and $J$ also cannot interchange. In other words, in Equation (13.69), there is no double count. All the states are independent.

But for non-massless states, this probably is not true anymore:

 $|$ $~\text{Number of states}$ ${a_1^I}^\dagger {a_1^J}^\dagger \bar a_1^{K\dagger} \bar a_1^{L\dagger} | p^+, \vec p_T \rangle~|$ $~\left[ \frac{(D-2)(D-1)}{2} \right]^2$ ${a_1^I}^\dagger {a_1^J}^\dagger \bar a_2^{K\dagger} | p^+, \vec p_T \rangle~|$ $~\left[ \frac{(D-2)(D-1)}{2} \right](D-2)$ ${a_2^I}^\dagger \bar a_1^{K\dagger} \bar a_1^{L\dagger} | p^+, \vec p_T \rangle~|$ $~(D-2)\left[ \frac{(D-2)(D-1)}{2} \right]$

— Me@2018-04-17 04:47:31 PM

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# Problem 14.4a1

Closed string degeneracies | A First Course in String Theory

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(a) State the values of $\alpha' M^2$ and give the degeneracies for the first five mass levels of the closed bosonic string theory.

~~~

What is the meaning of “closed bosonic string theory”?

Bosonic? Meaning commutating creation operators?

Closed?

String? Or superstring?

.

p.308 “We describe the position of classical bosonic strings using the string coordinates $X^\mu(\tau, \sigma)$. The $X^\mu(\tau, \sigma)$ are classical commuting variables: …

In quantum theory the $X^\mu$‘s become operators that do not generally commute.”

.

“To get fermions in string theory we introduce new dynamical world-sheet variables $\psi_1^\mu(\tau, \sigma)$ and $\psi_2^\mu(\tau, \sigma)$. The classical variables $\psi_\alpha^\mu$ ($\alpha = 1, 2$) are not ordinary commuting variables, but rather anticommuting variables.”

p.324 “In closed superstring theories spacetime bosons arise from the (NS, NS) sector and also from the (R, R) sector, since this sector is “doubly” fermionic. The spacetime fermions arise from the (NS, R) and (R, NS) sectors.”

.

type IIA

(NS+, NS+) (bosons), (NS+, R+) (fermions), (R-, NS+) (fermions), (R-, R+) (bosons)

type IIB

(NS+, NS+) (bosons), (NS+, R-) (fermions), (R-, NS+) (fermions), (R-, R-) (bosons)

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p.323 “Just like the massless states in bosonic closed string theory they carry two vector indices.”

This line implies that the phrase “bosonic closed string theory” does not refer to superstrings.

— Me@2018-04-03 10:00:13 AM

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Bosonic string theory is the original version of string theory, developed in the late 1960s. It is so called because it only contains bosons in the spectrum.

In the 1980s, supersymmetry was discovered in the context of string theory, and a new version of string theory called superstring theory (supersymmetric string theory) became the real focus. Nevertheless, bosonic string theory remains a very useful model to understand many general features of perturbative string theory, and many theoretical difficulties of superstrings can actually already be found in the context of bosonic strings.

— Wikipedia on Bosonic string theory

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What is the difference between closed string and closed superstring?

— Me@2018-04-09 09:41:18 PM

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‘Superstring theory’ is a shorthand for supersymmetric string theory because unlike bosonic string theory, it is the version of string theory that accounts for both fermions and bosons and incorporates supersymmetry to model gravity.

— Wikipedia on Superstring theory

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# M-theory is _not_ a string theory

These five theories have all been known since the middle of the 1980s. Some relationships between them were found soon after their discovery, but a clearer picture emerged only in the late 1990s. The limit of type II theory as the string coupling is taken to infinity was shown to give a theory in eleven dimensions. This theory is called M-theory, with the meaning of M to be decided when the nature of the theory becomes clear. It is known, however, the M-theory is _not_ a string theory. M-theory contains membranes (2-branes) and 5-branes, and these branes are not D-branes. M-theory may end up playing a prominent role in understanding string theory. The discovery of many other relationships between the five string theories listed above and M-theory has made it clear that we really have just _one theory_. This is a fundamental result: there is a unique theory, and the five superstrings and M-theory are different limits of this unique theory.

— Second Edition p.325

— A First Course in String Theory

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2018.04.05 Thursday ACHK

# Problem 14.4a

Closed string degeneracies.

For closed string states the left-moving and right-moving excitations are each described like states of open strings with identical values of $\alpha' M^2$. The value of $\alpha' M^2$ for the closed string state is four times that value.

How come?

.

Equation (14.78):

$\frac{1}{2} \alpha' M^2 = \alpha' M_L^2 + \alpha' M_R^2$

Mass M is two times that value ($M = 2 M_L$ or $M = 2 M_R$). So $M^2$ is four times.

— Me@2015.07.14 12:47 PM

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Not necessarily so.

— Me@2015.07.16 08:14 AM

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Instead, it is just due to this definition.

— Me@2015.07.30 09:19 AM

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p.322 “In the closed string theory the value of the mass-squared is given by…”

Equation (14.78) is actually a definition of the mass-squared of a closed string state.

Also, consider

p.322 “As befits closed strings there is also the level-matching condition $\alpha_0^- = \bar \alpha_0^-$ on the states. This condition guarantees that the left and right sectors give identical contributions to the mass-squared: $\alpha' M_L^2 = \alpha' M_R^2$.”

Then

$\frac{1}{2} \alpha' M^2 = \alpha' M_L^2 + \alpha' M_R^2$

$\alpha' M^2 = 2 \left( \alpha' M_L^2 + \alpha' M_R^2 \right) = 4 \alpha' M_L^2$

— Me@2018.03.02 11:05 AM

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# Problem 14.3b6

Quick Calculation 14.4b | A First Course in String Theory

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Massive level in the open superstring

Consider the first and second excited levels of the open superstring ($\alpha' M^2 = 1$ and $\alpha' M^2 = 2$). List the states in NS sector and the states in the R sector. Confirm that you get the same number of states.

~~~

When $\alpha' M^2 = 2$, by Equation (14.54), the possible states are

$\{ \alpha_{-2}^I, \alpha_{-1}^I \alpha_{-1}^J, d_{-1}^I d_{-1}^J \} | R_a \rangle, || ...$

$\{ \alpha_{-1}^I d_{-1}^J, d_{-2}^I \} | R_{\bar a} \rangle, || ...$

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For $\alpha_{-2}^I | R_a \rangle$, the number of states is 8.

For $\alpha_{-1}^I \alpha_{-1}^J | R_a \rangle$, the number of states is $\frac{8 \times 7}{2} + 8 = 36$.

For $d_{-1}^I d_{-1}^J | R_a \rangle$, the number of states is $\frac{8 \times 7}{2} = 28$.

For $\alpha_{-1}^I d_{-1}^J | R_{\bar a} \rangle$, the number of states is $8 \times 8 = 64$.

For $d_{-2}^I | R_{\bar a} \rangle$, the number of states is 8.

.

However, since each of $a$ and ${\bar a}$ has 8 possible values, there is an additional multiple of 8.

The total number of states is $8 \left[ 8 + 36 + 28 + 64 + 8 \right]$.

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You can check this answer against Equation (14.67):

$f_{NS} (x) = \frac{1}{\sqrt{x}} + 8 + 36 \sqrt{x} + 128 x + 402 x \sqrt{x} + 1152 x^2 + ...$

— Me@2018.02.20 10:57 AM

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# Problem 14.3b5

Quick Calculation 14.4b | A First Course in String Theory

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Massive level in the open superstring

Consider the first and second excited levels of the open superstring ($\alpha' M^2 = 1$ and $\alpha' M^2 = 2$). List the states in NS sector and the states in the R sector. Confirm that you get the same number of states.

~~~

When $\alpha' M^2 = 2$, by Equation (14.37),

$M^2 = \frac{1}{\alpha'} \left( - \frac{1}{2} + N^\perp \right)$

$N^\perp = \frac{5}{2}$

.

$\{ b_{-1/2}^I b_{-1/2}^J b_{-1/2}^K b_{-1/2}^L b_{-1/2}^M,$

$b_{-3/2}^I b_{-1/2}^L b_{-1/2}^M,$

$b_{-5/2}^I,$

$\alpha_{-1}^I b_{-1/2}^J b_{-1/2}^K b_{-1/2}^M,$

$\alpha_{-1}^I b_{-3/2}^J,$

$\alpha_{-2}^I b_{-1/2}^J,$

$\alpha_{-1}^I \alpha_{-1}^J b_{-1/2}^K \} |NS \rangle \otimes |p^+, \vec p_T \rangle$

.

There are $\frac{8 \times 7 \times 6 \times 5 \times 4}{5!} = {8 \choose 5} = 56$ states for

$\{ b_{-1/2}^I b_{-1/2}^J b_{-1/2}^K b_{-1/2}^L b_{-1/2}^M |NS \rangle \otimes |p^+, \vec p_T \rangle$

There are $8 \times \frac{8 \times 7}{2} = 224$ states for

$\{ b_{-3/2}^I b_{-1/2}^L b_{-1/2}^M \} |NS \rangle \otimes |p^+, \vec p_T \rangle$

There are $8$ states for

$\{ b_{-5/2}^I \} |NS \rangle \otimes |p^+, \vec p_T \rangle$

There are $8 \times {8 \choose 3} = 448$ states for

$\{ \alpha_{-1}^I b_{-1/2}^J b_{-1/2}^K b_{-1/2}^M \} |NS \rangle \otimes |p^+, \vec p_T \rangle$

There are $8 \times 8 = 64$ states for

$\{ \alpha_{-1}^I b_{-3/2}^J \} |NS \rangle \otimes |p^+, \vec p_T \rangle$

There are $8 \times 8 = 64$ states for

$\{ \alpha_{-2}^I b_{-1/2}^J \} |NS \rangle \otimes |p^+, \vec p_T \rangle$

There are $\left( \frac{8 \times 7}{2!} + 8 \right) \times 8 = 288$ states for

$\{ \alpha_{-1}^I \alpha_{-1}^J b_{-1/2}^K \} |NS \rangle \otimes |p^+, \vec p_T \rangle$

So the total number of states is 56 + 224 + 8 + 448 + 64 + 64 + 288 = 1152.

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You can check this answer against Equation (14.67):

$f_{NS} (x) = \frac{1}{\sqrt{x}} + 8 + 36 \sqrt{x} + 128 x + 402 x \sqrt{x} + 1152 x^2 + ...$

— Me@2018-02-16 03:22:13 PM

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# Ramond sector zero modes

Problem 14.3b4

A First Course in String Theory

What are $\xi_1, \xi_2, \xi_3, \xi_4$ in Equation (14.44)?

p.315 “Ramond fermions are more complicated than NS fermions because the eight fermionic zero mode $d_0^I$ must be treated with care. It turns out that these eight operators can be organized by simple linear combinations into four creation operators and four annihilation operators. Let us call the four creation operators …”

Since there are 8 possible transverse directions, there are 8 possible $d_0^I$‘s, where $I = 2,3, ..., 9$.

What is the meaning of “… organized by simple linear combinations into four creation operators …”?

— Me@2015.11.01 03:53 AM

The $d_0^I$ operators are similar to but different from other $d_r^I$ operators.

$d_0^I$‘s and $d_r^I$‘s are similar in the sense that they all follow Equation (14.43):

$\{ d_m^I, d_n^J \} = \delta_{m+n, 0} \delta^{IJ}$

p.315 “Again, the negatively moded oscillators $d_{-1}^I, d_{-2}^I, d_{-3}^I, ...$, are creation operators, while the positively moded ones $d_{1}^I, d_{2}^I, d_{3}^I, ...$ are annihilation operators.”

$d_0^I$‘s and $d_r^I$‘s are different in the sense that $d_0^I$‘s are neither creation nor annihilation operators.

(Based on the ideas from “Introduction to String Theory, A.N. Schellekens” and “A First Course in String Theory (Second Edition)” p.315:)

If we define $d_0 | 0 \rangle = 0$,

$\{ d_0^I, d_0^J \} | 0 \rangle$
$= \left( d_0^I d_0^J + d_0^I d_0^J \right) | 0 \rangle$
$= 0$

which does not match the requirement of

$\{ d_0^I, d_0^J \} = \delta^{IJ}$

So the definition $d_0 | 0 \rangle = 0$ does not work.

— Me@2015.11.12 11:30 AM

Instead, they are “organized by simple linear combinations into four creation operators”.

(Based on the idea from “Boundary Conformal Field Theory and the Worldsheet Approach to D-Branes, by Andreas Recknagel，Volker Schomerus” and “A First Course in String Theory (Second Edition)” p.315:)

Let

$c_0^i = d_0^{i+1}$
$e_i = \frac{1}{\sqrt{2}} \left( c_0^{2i} - i c_0^{2i - 1} \right)$
$e_i^\dagger = \frac{1}{\sqrt{2}} \left( c_0^{2i} + i c_0^{2i - 1} \right)$.

Then

$\left\{ e_i, e_j^\dagger \right\}$
$= \frac{1}{2} \left\{ \left( c_0^{2i} - i c_0^{2i - 1} \right), \left( c_0^{2j} + i c_0^{2j - 1} \right) \right\}$
$= \frac{1}{2} \delta^{ij} \left\{ \left( c_0^{2i} - i c_0^{2i - 1} \right), \left( c_0^{2i} + i c_0^{2i - 1} \right) \right\}$

By p.315 Equation (14.43):

$\{ d_0^I, d_0^J \} = \delta^{IJ}$

In other words,

$\{ c_0^{I-1}, c_0^{J-1} \} = \delta^{I-1,J-1}$
$\{ c_0^{I}, c_0^{J} \} = \delta^{IJ}$

$\left\{ e_i, e_j^\dagger \right\}$
$= \frac{1}{2} \left\{ \left( c_0^{2i} - i c_0^{2i - 1} \right), \left( c_0^{2j} + i c_0^{2j - 1} \right) \right\}$
$= \frac{1}{2} \delta^{ij} \left[\left\{ c_0^{2i} , c_0^{2i} \right\} - \left\{ i c_0^{2i - 1}, i c_0^{2i - 1} \right\} \right]$
$= \frac{1}{2} \delta^{ij} \left[\left\{ c_0^{2i} , c_0^{2i} \right\} + \left\{ c_0^{2i - 1}, c_0^{2i - 1} \right\} \right]$
$= \frac{1}{2} \delta^{ij} \left[1 + 1 \right]$
$= \delta^{ij}$

This is compatible with the anti-commutator requirement for fermion creation and annihilation operators:

$\{a^{\,}_i, a^\dagger_j\} = \delta_{i j}$

— Me@2015.11.13 11:14 PM

# Ground states and Annihilation operators

1. The equation $a | 0 \rangle = 0$ means that the eigenvalue of $a$ on $| 0 \rangle$ is 0:

$a | 0 \rangle = 0 | 0 \rangle$

2. The length of the vector $a | 0 \rangle$ is 0:

$\langle 0 | a^\dagger a | 0 \rangle = 0$

3. The physical meaning is that the probability of the system being at state $a | 0 \rangle$ is 0.

In other words, there is no state with an eigen-energy lower than the ground state one.

4. For the equation $a | 0 \rangle = 0 | 0 \rangle$, the 0 at the right is a scalar.

5. For the equation $a | 0 \rangle = 0$, the 0 at the right is a zero vector – a state vector with length zero.

6. $| 0 \rangle$ is a state vector. However, it is NOT the zero vector.

Instead, it is the state vector of the ground state. Its length is 1 unit.

— Me@2015-11-03 03:26:58 PM

# Problem 14.3b3

A First Course in String Theory

14.3 Massive level in the open superstring.

~~~

What is a zero mode?

p.315 “Being zero modes, these creation operators do not contribute to the mass-squared of the states.”

How come there are creation operators that do not contribute to the mass-squared of the states?

Before each creation operator, there is a multiple which is the same as the absolute value of the index of the creation operators.

If an index can be zero, the corresponding term can be zero.

— Me@2015.09.26 08:44 PM

Consider the NS-sector:

Equation (14.37):

$M^2 = \frac{1}{\alpha'} \left( \frac{-1}{2} + N^\perp \right)$

$N^\perp = \sum_{p=1}^\infty \alpha_{-p}^I \alpha_p^I + \sum_{r=\frac{1}{2}, \frac{3}{2} ...} r b_{-r}^I b_r^I$

For the NS sector, the $r$ values in $b^I_{r}$ are half-integers, thus cannot be zero. So every creation operator $b_{-r}^I$ contributes to the mass-squared.

(p.312 “… the negatively moded coefficients $b_{-1/2}^I$, $b_{-3/2}^I$, $b_{-5/2}^I$, …, are creation operators, …”)

Consider the R-sector:

Equation (14.53):

$M^2 = \frac{1}{\alpha'} \sum_{n \ge 1} \left( \alpha_{-n}^I \alpha_{n}^I + n d_{-n}^I d_n^I \right)$

For the R sector, the $n$ values in $d^I_n$ are integers, thus can be zero. So some of the creation operators $d^I_{-n}$ are zero modes.

p.315 “… the eight fermionic zero modes $d_0^I$…”

— Me@2015-10-11 11:01:44 AM